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Orthonormal-frame curvature components and the Kretschmann scalar 12s^2/r^6 of the Schwarzschild exterior metric

Statement

Let $(M, g, N)$ be the Schwarzschild exterior datum of mass parameter $m_{*}$ , Schwarzschild radius $s = 2 G_{N} m_{*} / c^{2}$ and $f (r) = 1 - s / r$ (The Schwarzschild exterior model datum: domain, metric, mass parameter and time orientation), with curvature data as in Curvature components and the vanishing of the Ricci and Einstein tensors of the Schwarzschild exterior metric. On each spherical-polar product chart $(x^{0}, r, ϑ, φ)$ define the local frame field

$e_{\hat{0}} : = f^{- 1 / 2} \partial_{0}, e_{\hat{1}} : = f^{1 / 2} \partial_{1}, e_{\hat{2}} : = r^{- 1} \partial_{2}, e_{\hat{3}} : = (r \sin ϑ)^{- 1} \partial_{3},$

and for the curvature tensor define the frame components

${\hat{R}}_{\hat{a} \hat{b} \hat{c} \hat{d}} : = g (R (e_{\hat{c}}, e_{\hat{d}}) e_{\hat{b}}, e_{\hat{a}}) .$

Then:

(a) The frame field. $(e_{\hat{0}}, e_{\hat{1}}, e_{\hat{2}}, e_{\hat{3}})$ is a smooth $g$ -orthonormal frame (tetrad field) on the chart domain, with $g (e_{\hat{a}}, e_{\hat{b}}) = diag (- 1, 1, 1, 1)_{\hat{a} \hat{b}}$ , and $e_{\hat{0}}$ is future-directed timelike; it is a tetrad field in the sense of Local frame fields, coordinate frames and tetrad fields. It is not a coordinate frame.

(b) Frame components. The nonzero frame components, up to the symmetries $R_{a b c d} = - R_{b a c d} = - R_{a b d c} = R_{c d a b}$ of The algebraic symmetries of the curvature tensor and the symmetry of the Ricci and Einstein tensors, are

${\hat{R}}_{\hat{0} \hat{1} \hat{0} \hat{1}} = \frac{f^{''}}{2}, {\hat{R}}_{\hat{0} \hat{2} \hat{0} \hat{2}} = {\hat{R}}_{\hat{0} \hat{3} \hat{0} \hat{3}} = \frac{f^{'}}{2 r},$

${\hat{R}}_{\hat{1} \hat{2} \hat{1} \hat{2}} = {\hat{R}}_{\hat{1} \hat{3} \hat{1} \hat{3}} = - \frac{f^{'}}{2 r}, {\hat{R}}_{\hat{2} \hat{3} \hat{2} \hat{3}} = \frac{1 - f}{r^{2}},$

with $f^{'} = s / r^{2}$ and $f^{''} = - 2 s / r^{3}$ ; all other frame components vanish.

(c) Kretschmann scalar. Define the global smooth scalar field $K : = R_{a b c d} R^{a b c d}$ , with indices raised by the inverse metric. On the spherical-polar chart locus its component expression is

$K = f^{'' 2} + \frac{4 (f^{'})^{2}}{r^{2}} + \frac{4 (1 - f)^{2}}{r^{4}},$

and for the model function $f = 1 - s / r$ this is

$K = \frac{12 s^{2}}{r^{6}} = \frac{48 G_{N}^{2} m_{*}^{2}}{c^{4} r^{6}} .$

The scalar identity is first computed on the regular angular locus and holds on all of $M$ by smoothness and density. Since $r > s$ on $M$ , one has $0 < K = 12 s^{2} / r^{6} < 12 / s^{4}$ at every point; as $r ↓ s$ the expression has the finite boundary value $12 / s^{4}$ , and the expression $12 s^{2} / r^{6}$ , read as a function of $r > 0$ , grows without bound as $r ↓ 0$ .

(d) Scope. The lemma computes curvature data of the exterior datum on $r > s$ . The limit $r ↓ s$ is the boundary value on $M$ of the smooth expression $12 s^{2} / r^{6}$ ; no value of the metric at $r = s$ or at $r = 0$ is used or asserted here, those regions being outside the datum (the regular continuation across $r = s$ is the separate extension lemma of this page).

Facts & Assumptions

Given: The Schwarzschild exterior datum, a spherical-polar product chart $(x^{0}, r, ϑ, φ)$ with $0 < ϑ < π$ , the functions $f = 1 - s / r$ , $f^{'} = s / r^{2}$ , $f^{''} = - 2 s / r^{3}$ , and the local frame field $e_{\hat{a}}$ of the statement.

[L1]

The metric components and inverse components in the chart are $g_{00} = - f$ , $g_{11} = f^{- 1}$ , $g_{22} = r^{2}$ , $g_{33} = r^{2} \sin^{2} ϑ$ , $g^{00} = - f^{- 1}$ , $g^{11} = f$ , $g^{22} = r^{- 2}$ , $g^{33} = r^{- 2} \sin^{- 2} ϑ$ , off-diagonal components zero (The Schwarzschild exterior datum is a smooth time-oriented Lorentzian four-manifold, The inverse metric, musical isomorphisms and index raising).

[L2]

The nonzero lowered curvature components in the chart are $R_{0101} = f^{''} / 2$ , $R_{0202} = r f f^{'} / 2$ , $R_{0303} = r f f^{'} \sin^{2} ϑ / 2$ , $R_{1212} = - r f^{'} / (2 f)$ , $R_{1313} = - r f^{'} \sin^{2} ϑ / (2 f)$ , $R_{2323} = r^{2} (1 - f) \sin^{2} ϑ$ , and every component whose index pairs do not coincide as unordered pairs, or with a repeated index inside a pair, vanishes (Curvature components and the vanishing of the Ricci and Einstein tensors of the Schwarzschild exterior metric clauses (a) and (d)).

[L3]

Curvature components transform as those of a $(0, 4)$ -tensor: if $e_{\hat{a}} = d_{\hat{a}}^{μ} \partial_{μ}$ are the coefficients of a frame field, then the frame components of the lowered curvature are ${\hat{R}}_{\hat{a} \hat{b} \hat{c} \hat{d}} = d_{\hat{a}}^{α} d_{\hat{b}}^{β} d_{\hat{c}}^{γ} d_{\hat{d}}^{δ} R_{α β γ δ}$ evaluated on the same chart domain; the tensor $R$ is smooth, its components transform by this law, and tensor identities and scalar contractions are intrinsic (Tensoriality of the curvature, its coordinate formula and its point-normal metric formula clauses (a)-(b), Tensor fields of type (r,s) on a Lorentzian four-manifold and their components, The algebraic symmetries of the curvature tensor and the symmetry of the Ricci and Einstein tensors).

[L4]

The scalar $R_{a b c d} R^{a b c d}$ with $R^{a b c d} = g^{a α} g^{b β} g^{c γ} g^{d δ} R_{α β γ δ}$ is frame independent. In an orthonormal Lorentz frame each ordered component contributes $(- 1)^{m} ({\hat{R}}_{\hat{a} \hat{b} \hat{c} \hat{d}})^{2}$ , where $m$ is the number of time indices among $\hat{a}, \hat{b}, \hat{c}, \hat{d}$ . The only nonzero frame components listed in clause (b), and all their symmetry variants, have either zero or two time indices, so every nonzero contribution has positive sign (The inverse metric, musical isomorphisms and index raising, Tensor fields of type (r,s) on a Lorentzian four-manifold and their components, Local frame fields, coordinate frames and tetrad fields, Curvature components and the vanishing of the Ricci and Einstein tensors of the Schwarzschild exterior metric clause (a)).

[L5]

The model functions satisfy $f^{'} = s / r^{2}$ , $f^{''} = - 2 s / r^{3}$ ; on the chart domain $r > s > 0$ and $\sin ϑ > 0$ , and the displayed expressions are smooth and positive or negative as indicated (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Rational powers $a^{r}$ of a positive base).

Proof

technique · verify the frame is orthonormal, rescale the coordinate curvature components of the preceding lemma into the frame, and add the squares of the frame components

1.1L1L5givenalgebra

(Orthonormality.) By [L1], $g (e_{\hat{0}}, e_{\hat{0}}) = f^{- 1} g_{00} = - 1$ , $g (e_{\hat{1}}, e_{\hat{1}}) = f g_{11} = 1$ , $g (e_{\hat{2}}, e_{\hat{2}}) = r^{- 2} g_{22} = 1$ and $g (e_{\hat{3}}, e_{\hat{3}}) = (r \sin ϑ)^{- 2} g_{33} = 1$ . Each $e_{\hat{a}}$ is a smooth multiple of a coordinate field by a smooth nonvanishing function on the chart domain, so the $e_{\hat{a}}$ are smooth vector fields and form a basis at every point; the mixed pairings vanish because the coordinate frame is $g$ -orthogonal by [L1]. Hence the frame is $g$ -orthonormal with the stated diagonal pairing. Since $e_{\hat{0}} = f^{- 1 / 2} \partial_{0}$ with $f^{- 1 / 2} > 0$ and $\partial_{0}$ is future-directed by the time orientation of the datum, $e_{\hat{0}}$ is future-directed. This proves (a).

2.1L2L3step 1.1algebra

(Frame components.) The coefficient matrix $d_{\hat{a}}^{μ}$ of [L3] is diagonal with entries $d_{\hat{0}}^{0} = f^{- 1 / 2}$ , $d_{\hat{1}}^{1} = f^{1 / 2}$ , $d_{\hat{2}}^{2} = r^{- 1}$ , $d_{\hat{3}}^{3} = (r \sin ϑ)^{- 1}$ . Applying [L3] to the nonzero components of [L2]: ${\hat{R}}_{\hat{0} \hat{1} \hat{0} \hat{1}} = R_{0101} (d_{\hat{0}}^{0})^{2} (d_{\hat{1}}^{1})^{2} = (f^{''} / 2) f^{- 1} f = f^{''} / 2$ ; ${\hat{R}}_{\hat{0} \hat{2} \hat{0} \hat{2}} = R_{0202} (d_{\hat{0}}^{0})^{2} (d_{\hat{2}}^{2})^{2} = (r f f^{'} / 2) f^{- 1} r^{- 2} = f^{'} / (2 r)$ ; ${\hat{R}}_{\hat{0} \hat{3} \hat{0} \hat{3}} = R_{0303} (d_{\hat{0}}^{0})^{2} (d_{\hat{3}}^{3})^{2} = (r f f^{'} \sin^{2} ϑ / 2) f^{- 1} (r \sin ϑ)^{- 2} = f^{'} / (2 r)$ ; ${\hat{R}}_{\hat{1} \hat{2} \hat{1} \hat{2}} = R_{1212} (d_{\hat{1}}^{1})^{2} (d_{\hat{2}}^{2})^{2} = (- r f^{'} / (2 f)) f r^{- 2} = - f^{'} / (2 r)$ ; ${\hat{R}}_{\hat{1} \hat{3} \hat{1} \hat{3}} = R_{1313} (d_{\hat{1}}^{1})^{2} (d_{\hat{3}}^{3})^{2} = (- r f^{'} \sin^{2} ϑ / (2 f)) f (r \sin ϑ)^{- 2} = - f^{'} / (2 r)$ ; and ${\hat{R}}_{\hat{2} \hat{3} \hat{2} \hat{3}} = R_{2323} (d_{\hat{2}}^{2})^{2} (d_{\hat{3}}^{3})^{2} = r^{2} (1 - f) \sin^{2} ϑ r^{- 2} (r \sin ϑ)^{- 2} = (1 - f) / r^{2}$ . Every other frame component is a diagonal coefficient multiple of a coordinate component that vanishes by [L2], hence vanishes. This proves (b) and identifies all nonzero frame components.

3.1L2L3L4step 2.1algebra

(The Kretschmann scalar from the frame components.) By [L4] and the fact that every nonzero component has zero or two time indices, each nonzero ordered quadruple contributes the positive square of its component. The pair symmetries of [L2] and [L3] show that for each of the six unordered pairs $P$ of distinct frame indices, the four ordered quadruples realizing $P$ in both pairs contribute four times the square of the corresponding component of step 2.1. Hence $K = 4 (f^{''} / 2)^{2} + 4 (f^{'} / (2 r))^{2} \cdot 4 + 4 ((1 - f) / r^{2})^{2}$ : the first term is the pair ${0, 1}$ ; the factor $4 \cdot 4$ collects the four pairs ${0, 2}, {0, 3}, {1, 2}, {1, 3}$ each contributing $4 (f^{'} / (2 r))^{2}$ ; and the last term is the pair ${2, 3}$ . Simplifying, $K = f^{'' 2} + 4 (f^{'})^{2} / r^{2} + 4 (1 - f)^{2} / r^{4}$ . The contraction is a smooth scalar on all of $M$ , and this expression is smooth there; they agree on the dense regular angular locus, so they agree at the poles by continuity. This proves (c) up to the model specialisation.

4.1L5step 3.1algebra∎

(Specialisation and limits.) For $f = 1 - s / r$ one has $f^{'} = s / r^{2}$ and $f^{''} = - 2 s / r^{3}$ , so $f^{'' 2} = 4 s^{2} / r^{6}$ , $4 (f^{'})^{2} / r^{2} = 4 s^{2} / r^{6}$ and $4 (1 - f)^{2} / r^{4} = 4 s^{2} / r^{6}$ ; the sum is $12 s^{2} / r^{6} = 48 G_{N}^{2} m_{*}^{2} / (c^{4} r^{6})$ , using $s = 2 G_{N} m_{*} / c^{2}$ . On $M$ the identity $K = 12 s^{2} / r^{6}$ holds at every point with $r > s > 0$ , so $0 < K < 12 / s^{4}$ ; as $r ↓ s$ the right-hand side tends to the finite value $12 / s^{4}$ , as $r \to + \infty$ it tends to $0$ , and as a function of $r > 0$ the expression $12 s^{2} / r^{6}$ tends to $+ \infty$ as $r ↓ 0$ . These are statements about the displayed expression, not about the metric at $r = s$ or $r = 0$ ; no value of the metric outside $M$ is used or asserted, which is clause (d).

Remarks

Conventions and the source value. In the literature the invariant is often written $48 G^{2} M^{2} / r^{6}$ in units $G = c = 1$ ; with our $s = 2 G_{N} m_{*} / c^{2}$ this is the same number as $12 s^{2} / r^{6}$ . The displayed value $12 G^{2} M^{2} / r^{6}$ appearing in one source cited above is a factor-4 discrepancy from the direct computation performed here and is not inherited; the other cited source gives the consistent value.
What the invariant shows. The boundary value $12 / s^{4}$ of $K$ at $r ↓ s$ shows that the divergence of the chart component $g_{11}$ at $r = s$ is not accompanied by a divergence of this curvature invariant; on the smooth extension of the next lemma, where $r$ ranges over $(0, \infty)$ and $K = 12 s^{2} / r^{6}$ holds there, the same expression grows without bound as $r ↓ 0$ . Both facts are used in the horizon-and-singularity remark of this page.

Orthonormal-frame curvature components and the Kretschmann scalar 12s^2/r^6 of the Schwarzschild exterior metric

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources